AQA A-Level Chemistry: Rate equations (7405)

This guide teaches section 3.1.9 Rate equations of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for exams from June 2016 onwards: 3.1.9.1, 3.1.9.2 and Required practical 7. The whole section is A-level only. The specification places it on A-level Paper 2, and Paper 3 can test any content.

For quick recall, use the rate equations revision notes. To test yourself, use the rate equations practice questions. It builds on kinetics (3.1.5). The course hub is AQA A-Level Chemistry, and the printable checklist lists every specification point.

What this unit covers

Spec point	What you must be able to do	Status
3.1.9.1	Define order and rate constant; calculate with the rate equation	A-level only
3.1.9.1	Effect of temperature on k; calculate with k = Ae^(−Ea/RT); plot ln k against 1/T (slope −Ea/R)	A-level only
3.1.9.2	Rates and initial rates from concentration–time graphs; k of a zero-order reaction from a gradient	A-level only
3.1.9.2	Orders (0, 1, 2) from rate–concentration data or graphs; derive the rate equation; link orders to the rate-determining step	A-level only
Required practical 7	Measure rate by an initial rate method and a continuous monitoring method	A-level only

3.1.9.1 The rate equation

For a reaction involving reactants A and B, the rate is linked to their concentrations by a rate equation:

Rate = k[A]^m [B]^n

[A] is concentration in mol dm⁻³; rate is in mol dm⁻³ s⁻¹.

• Order of reaction with respect to a reactant: the power to which the concentration of that reactant is raised in the rate equation. Here, m is the order with respect to A and n the order with respect to B.
• Overall order: the sum of the individual orders, m + n.
• Rate constant, k: the constant of proportionality in the rate equation. It is fixed for a given reaction at a given temperature.

Orders are restricted to 0, 1 and 2:

Order with respect to A	Doubling [A] multiplies rate by	Tripling [A] multiplies rate by
0	1	1
1	2	3
2	4	9

A zero-order reactant is still used up, but it does not appear in the rate equation ([A]⁰ = 1). Orders come from experiments, not from the balanced equation.

Units of k

Rearrange for k, put in the units and cancel.

Overall order	Units of k
0	mol dm⁻³ s⁻¹
1	s⁻¹
2	mol⁻¹ dm³ s⁻¹
3	mol⁻² dm⁶ s⁻¹

Worked example 1 (rate from a rate equation). For a reaction, Rate = k[A][B]² and k = 3.6 mol⁻² dm⁶ s⁻¹ at 298 K. Calculate the rate when [A] = 0.050 mol dm⁻³ and [B] = 0.20 mol dm⁻³. Then state the factor by which the rate changes if [A] is tripled and [B] is halved.

Rate = 3.6 × 0.050 × (0.20)²
     = 3.6 × 0.050 × 0.040
     = 7.2 × 10⁻³ mol dm⁻³ s⁻¹
Factor = 3 × (½)² = 3 × ¼ = 0.75

Worked example 2 (k and its units). For Rate = k[C]², the rate is 4.5 × 10⁻⁴ mol dm⁻³ s⁻¹ when [C] = 0.030 mol dm⁻³. Calculate k.

k = Rate ÷ [C]² = 4.5 × 10⁻⁴ ÷ (0.030)² = 4.5 × 10⁻⁴ ÷ 9.0 × 10⁻⁴ = 0.50
units = (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)² = mol⁻¹ dm³ s⁻¹
k = 0.50 mol⁻¹ dm³ s⁻¹

3.1.9.1 Temperature and the rate constant

Concentration changes the rate but not k. When the temperature rises, k increases, so the rate rises at the same concentrations. A small rise gives a large increase in k because a much larger proportion of collisions have energy at or above the activation energy, Ea (see 3.1.5).

The Arrhenius equation:

k = A e^(−Ea/RT)

• A is a constant, the Arrhenius constant. It has the same units as k.
• Ea is the activation energy, in J mol⁻¹ in this equation.
• R is the gas constant, 8.31 J K⁻¹ mol⁻¹.
• T is the temperature in K.

The specification states that these equations and R will be given when required.

Worked example 3 (Arrhenius calculation). A first-order reaction has A = 4.0 × 10¹⁰ s⁻¹ and Ea = 75 kJ mol⁻¹. Calculate k at 310 K and at 320 K.

Ea = 75 × 1000 = 75 000 J mol⁻¹
At 310 K: Ea/RT = 75 000 ÷ (8.31 × 310) = 29.11
          k = 4.0 × 10¹⁰ × e^(−29.11) = 9.08 × 10⁻³ s⁻¹
At 320 K: k = 4.0 × 10¹⁰ × e^(−75 000 / (8.31 × 320)) = 2.26 × 10⁻² s⁻¹

A 10 K rise multiplies k by about 2.5 here.

The straight-line form

Taking natural logs of both sides gives:

ln k = −Ea/(RT) + ln A

This has the form y = mx + c, with y = ln k, x = 1/T, gradient m = −Ea/R and intercept c = ln A. So a plot of ln k against 1/T is a straight line with slope −Ea/R.

Worked example 4 (Ea from a graph). A first-order reaction gives these values.

T / K	k / s⁻¹	1/T / K⁻¹	ln k
300	0.0316	3.333 × 10⁻³	−3.45
320	0.150	3.125 × 10⁻³	−1.90
340	0.590	2.941 × 10⁻³	−0.53
360	2.00	2.778 × 10⁻³	0.69

Step 1: calculate 1/T and ln k (shown) and plot ln k against 1/T; the points lie on a straight line. Step 2: take the gradient from two points far apart:

gradient = (0.69 − (−3.45)) ÷ (2.778 × 10⁻³ − 3.333 × 10⁻³)
         = 4.14 ÷ (−5.55 × 10⁻⁴) = −7.46 × 10³ K

Step 3: Ea = −gradient × R = 7.46 × 10³ × 8.31 = 6.20 × 10⁴ J mol⁻¹ = 62.0 kJ mol⁻¹.

Keep 1/T to at least four significant figures, or the small differences lose accuracy.

3.1.9.2 Rates from concentration–time graphs

The rate at any moment is the size of the gradient of the concentration–time graph then. For a curve, draw a tangent and find its gradient (negative for a reactant; quote the rate as positive).

The initial rate is the gradient of the tangent drawn at t = 0.

Worked example 5 (tangent). On a concentration–time curve for reactant D, the tangent at t = 60 s passes through (0 s, 0.62 mol dm⁻³) and (150 s, 0.11 mol dm⁻³).

gradient = (0.11 − 0.62) ÷ (150 − 0) = −3.4 × 10⁻³ mol dm⁻³ s⁻¹
rate at 60 s = 3.4 × 10⁻³ mol dm⁻³ s⁻¹

Zero-order reactions

If the reaction is zero order with respect to X, Rate = k, so the rate stays constant as X is used up. The concentration–time graph is a straight line, and k = −gradient.

Worked example 6. [X] falls from 0.80 mol dm⁻³ at t = 0 to 0.44 mol dm⁻³ at t = 300 s, along a straight line.

gradient = (0.44 − 0.80) ÷ 300 = −1.2 × 10⁻³ mol dm⁻³ s⁻¹
k = 1.2 × 10⁻³ mol dm⁻³ s⁻¹

For first and second order the graph is a curve that flattens as the concentration falls.

3.1.9.2 Deducing orders

From rate–concentration graphs

Order	Rate against [A]
0	horizontal straight line
1	straight line through the origin
2	upward curve through the origin; rate against [A]² is a straight line through the origin

From initial rates data

Compare two experiments in which only one concentration changes. If two change together, divide out the effect of the one whose order you know.

Worked example 7. Data for D + E + F → products at a fixed temperature.

Exp	[D] / mol dm⁻³	[E] / mol dm⁻³	[F] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.10	0.10	0.20	2.0 × 10⁻⁴
2	0.20	0.10	0.20	8.0 × 10⁻⁴
3	0.10	0.30	0.20	6.0 × 10⁻⁴
4	0.10	0.10	0.40	2.0 × 10⁻⁴

Step 1, D (Exps 1 and 2): [D] doubles, rate ×4, so second order in D. Step 2, E (Exps 1 and 3): [E] triples, rate ×3, so first order in E. Step 3, F (Exps 1 and 4): [F] doubles, rate unchanged, so zero order in F. Step 4, rate equation: Rate = k[D]²[E] (overall order 3). Step 5, k from Exp 1:

k = 2.0 × 10⁻⁴ ÷ ((0.10)² × 0.10) = 0.20 mol⁻² dm⁶ s⁻¹

Check: Exp 2 gives 0.20 × (0.20)² × 0.10 = 8.0 × 10⁻⁴. It matches.

3.1.9.2 Orders and the rate-determining step

In a multi-step reaction the slowest step is the rate-determining (rate-limiting) step.

• A reactant in the rate equation takes part in the rate-determining step (or a step before it).
• Its order tells you how many particles of it take part up to and including that step.
• A zero-order reactant takes part only after the rate-determining step.

Worked example 8. The reaction 2P + Q → P₂Q has Rate = k[P][Q]. Which of these mechanisms fits?

• Mechanism I: P + Q → PQ (slow), then PQ + P → P₂Q (fast)
• Mechanism II: P + P → P₂ (slow), then P₂ + Q → P₂Q (fast)

Step 1: the rate equation shows one P and one Q in the rate-determining step. Step 2: the slow step of I uses one P and one Q, and the steps add up to the overall equation, so I fits. Step 3: the slow step of II would give Rate = k[P]², so II does not fit.

Required practical 7: measuring rate

The specification requires you to measure the rate of reaction by an initial rate method and by a continuous monitoring method (apparatus and techniques AT a, k and l). It suggests finding the order for a reactant in the iodine clock reaction.

Initial rate method (clock reaction). In an iodine clock, iodine forms slowly and reacts at once with a small, fixed amount of thiosulfate ions. When the thiosulfate runs out, iodine stays in solution and turns starch blue-black.

1. Measure solutions with a pipette or burette. Vary the volume of one reactant only, and add water so the total volume is the same each run; its concentration is then proportional to its volume.
2. Mix, start the timer, and stop it at the colour change. Keep the temperature constant.

The same small amount of product forms each run, so initial rate ∝ 1/t. Plot 1/t against the varied concentration (or volume) and read the order from the shape, as in the table above.

Continuous monitoring method. Follow one concentration through a single run: gas volume, mass loss, colour with a colorimeter, or samples removed, quenched (for example in ice-cold water) and titrated. Plot concentration against time and draw tangents to find rates at different concentrations.

Wear eye protection and handle each chemical according to its hazards (AT k).

Common errors

• Deducing orders from the coefficients in the balanced equation.
• Writing units of k from memory instead of cancelling them for the overall order.
• Using Ea in kJ mol⁻¹ or T in °C in k = Ae^(−Ea/RT).
• Sign errors: Ea = −gradient × R.
• Crediting a rate change to one reactant when two concentrations changed.
• Saying a zero-order reactant is not used up. It is; it reacts after the rate-determining step.

Next steps

Next, use the revision notes and the practice questions. The exam preparation guide covers paper structure, and the free 10-minute diagnostics show where to focus.

Official syllabus

AQA, AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: section 3.1.9 Rate equations (3.1.9.1 and 3.1.9.2, A-level only), with Required practical 7.